A rod of length 10 cm carries a charge of +50 uniformly distributed...

Length of the rod, $L = 10 \text{ cm} = 0.1 \text{ m}$

Charge on the rod, $Q = 50 \mu\text{C} = 50 \times 10^{-6} \text{ C}$

Linear charge density, $\lambda = \dfrac{Q}{L} = \dfrac{50 \times 10^{-6}}{0.1} = 5 \times 10^{-4} \text{ C/m}$

The point is at a distance of $10 \text{ cm}$ from both ends of the rod. Thus, the point and the two ends of the rod form an equilateral triangle of side $10 \text{ cm}$. The point lies on the perpendicular bisector of the rod.

The perpendicular distance of the point from the rod is $r = \sqrt{10^2 - 5^2} = 5\sqrt{3} \text{ cm} = 5\sqrt{3} \times 10^{-2} \text{ m}$.

The angle subtended by each half of the rod at the point is \theta_1 = \theta_2 = 30^{}^{{}^{\circ}}.

The electric field at the point is given by:

$E = \dfrac{k \lambda}{r} (\sin \theta_1 + \sin \theta_2)$

Substituting the values:

E = \dfrac{9 \times 10^9 \times 5 \times 10^{-4}}{5\sqrt{3} \times 10^{-2}} (\sin 30^{}^{{}^{\circ}} + \sin 30^{}^{{}^{\circ}})

$E = \dfrac{45 \times 10^5}{5\sqrt{3} \times 10^{-2}} \times 1$

$E = \dfrac{9 \times 10^7}{\sqrt{3}} = 3\sqrt{3} \times 10^7 \text{ N/C}$

$E \approx 3 \times 1.732 \times 10^7 \text{ N/C} = 5.196 \times 10^7 \text{ N/C} \approx 5.2 \times 10^7 \text{ N/C}$